(New page: 1. This signal is the math equivalent of a synonym for the word 'strange'. 2. This signal is never less than 0, but is always less than 8. 3. This signal's a_k is always 0 when it's grea...) |
(→answer) |
||
| Line 16: | Line 16: | ||
From 3, we know that every a_k not equal to 0, -1, or 1 will be equal to 0... this verifies that it's a sinewave. | From 3, we know that every a_k not equal to 0, -1, or 1 will be equal to 0... this verifies that it's a sinewave. | ||
| − | From 4, we know that the period is <math>\frac{2\pi}{5}</math> | + | From 4, we know that the period is 5, making <math>\omega=\frac{2\pi}{5}</math> |
Thus this signal is <math>4sin(\frac{2\pi}{5}t)+4</math> | Thus this signal is <math>4sin(\frac{2\pi}{5}t)+4</math> | ||
Revision as of 06:41, 25 September 2008
1. This signal is the math equivalent of a synonym for the word 'strange'.
2. This signal is never less than 0, but is always less than 8.
3. This signal's a_k is always 0 when it's greater than the absolute value of 1.
4. This signals period and the number of fingers on your hand share a common integer.
answer
From 1, we know that the signal is odd... a sinewave is probably a safe guess.
From 2, we know that the signal is bounded by 0 and 8.. this means it has an amplitude of 4 and an offset of 4.
From 3, we know that every a_k not equal to 0, -1, or 1 will be equal to 0... this verifies that it's a sinewave.
From 4, we know that the period is 5, making $ \omega=\frac{2\pi}{5} $
Thus this signal is $ 4sin(\frac{2\pi}{5}t)+4 $
